Question

Cho mọi số tự nhiên n\(\ge\)2

A=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\) với \(\frac{1}{2}\)

Trình bayd rõ ràng

Answer 1

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\right)\)

\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(A< \frac{1}{4}.\left(2-\frac{1}{n}\right)\)

\(A< \frac{1}{4}.2=\frac{1}{2}\left(đpcm\right)\)

Answer 2

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(\Rightarrow A=\frac{1}{\left(1.2\right)^2}+\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+....+\frac{1}{\left(2n\right)^2}\)

\(\Rightarrow A=\frac{1}{1^2.2^2}+\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+...+\frac{1}{2^2.n^2}\)

\(\Rightarrow A=\frac{1}{1}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+...+\frac{1}{2^2}.\frac{1}{n^2}\)

\(\Rightarrow A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2^2}+\frac{1}{n^2}\right)\)

Có: \(1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2^2}+\frac{1}{n^2}\) > 1

Rồi bạn tự tính tiếp nhé.